Gas Law Stoichiometry BHS-PS1-9 Level 4

= V1n2 = V2n1 V1 n1 V2 n2 n2 n1 n1 n2 V2= V1 V1= V2 When to use Gas Law Stoichiometry…. 1. If you have the volume of one gas and want the volume of a different gas (temperature & pressure do not matter) * use Avogadro’s Law (volume is directly proportional to moles) * V1 n1 V2 n2 = V1n2 = V2n1 n2 n1 n1 n2 V2= V1 V1= V2

Sample problem If 2.30 L N2 reacts at some temperature and pressure, what volume of NH3 forms? N2(g) + 3H2(g) → 2NH3(g) n2 n1 V2= V1 2.30 L N2 1 X 2 mol NH3 1 mol N2 =4.60 L NH3 NOTICE!! When you are calculating volume and changing the gas molecule all you need to do is use the MOLE RATIO

2. Involves using the volume/mass of a gas to calculate the mass/volume of another substance when temperature and pressure are at STP * at STP…. 1 mol = 22.4 L * use stoichiometry to convert volume/mass to mass/volume of other substance We did this in Unit 1!!

4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) Sample problem How many grams of NO will be produced when 26.8 L of water vapor are produced at STP? 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) 26.8 L H2O 1 X 1 mol H2O 22.4 L H2O X 4 mol NO 6 mol H2O 30.01 g NO 1 mol NO X =23.9 g NO

3. Involves using the volume/mass of a gas to calculate the mass/volume of another substance when temperature and pressure are NOT at STP volume A  mass B * use Ideal Gas Law (PV=nRT) to solve for n (moles) * use stoichiometry to convert moles to mass of other substance mass A  volume B * use stoichiometry to change the substance and convert to moles * use Ideal Gas Law (PV=nRT) to solve for volume

P V = n R T 3 H2 + N2  2 NH3 PV RT n= Sample problem A chemist might commonly perform this reaction (Haber process) in a chamber at 327oC under a pressure of 900. mm Hg. How many grams of ammonia would be produced from 166.3 liters of hydrogen at the above conditions? 3 H2 + N2  2 NH3 P V = n R T P = 900. mmHg V = 166.3 L H2 n = ? R =62.4 T = 327°C + 273 = 600 K mmHg L mol K PV RT n=

PV RT n= 3 H2 + N2  2 NH3 (900.mmHg)(166.3L H2) 600 K First - use PV=nRT to solve for moles (900.mmHg)(166.3L H2) 600 K n= X mol K 62.4 mmHg L =4.00 mol H2 Second - use stoich to change molecule and solve for grams 4.00 mol H2 1 X 2 mol NH3 3 mol H2 17.034 g 1 mol X =45.4 g NH3

P V = n R T nRT P V= Sample problem If magnesium is added to hydrochloric acid (HCl), hydrogen gas and magnesium chloride are made. If 3.46g of HCl is used, what volume of hydrogen gas will be produced at 1.40 atm and 25.0°C? Mg(s) + 2HCl(aq)  H2(g) + MgCl2(aq) P V = n R T P = 1.40 atm V = ? n = ? R = 0.0821 T = 25.0°C + 273 = 298 K atm L mol K nRT P V=

nRT P V= (0.04745 mol H2)(298 K) 1.40 atm 0.0821 atm L mol K V= Mg(s) + 2HCl(aq)  H2(g) + MgCl2(aq) First - use stoich to change molecule and solve for moles (n) 3.46 g HCl 1 1 mol HCl 36.458 g HCl X X 1 mol H2 2 mol HCl =0.04745 mol H2 Second - use PV=nRT to solve for volume (0.04745 mol H2)(298 K) 1.40 atm V= X 0.0821 atm L mol K =0.829 L H2 =829 mL H2